I need to know how to find the two answers between 0 and 2 pi. I need the answers in degrees, not radians. there should be two answers

sin x = (e^ix – e^-ix)/2i

Let sin x = -4 + sqrt2/2 = k = (e^ix – e^-ix)/2i

2ik = (e^ix – e^-ix)

(e^ix – e^-ix) –2ik = 0

e^2ix – 2ike^ix –1 = 0

Let u = e^ix

u^2 – 2iku – 1 = 0, u = (2ik ± (((-2ik)^2 – 4(1)(-1))^(1/2))/2

(-2.586i ± 1.64)/2

u= e^ix ≈ -.4735i, -2.113i

e^ix = -.4735i

ix = ln(-.4735i)

ix = ln(.4735) - (pi)/2

x = -i ln(.4735) – (pi)/2

x = -i(-.748) + (pi)/2

x = -1.57 + .748i

|x| = 1.739, Θ = arctan (.748/-1.57) = 154.5° OR 334.5º

AND

e^ix = -2.113i

ix = ln (-2.113i)

ix = ln (2.113) – ipi/2

x = -i ln (2.113) + (pi)/2

x = -1.57 - .748i

|x| = 1.739, Θ = 25.5° OR 205.5°

## Comments

Hi Abby.

There is huge difference between -4 + sin(x) = -8 + v2/2 andYou have to understand the order of operations!-4 + sin(x) = (-8 + v2) / 2 !!!!!!!

The answer for first one is

, and for another one:empty set of numberssin(x) = -4 + v2/2 + 4

sin(x) = v2/2 ---> x

_{1}= 45^{o}and x_{2}= 135^{o}on the segment [0, 2pi].